Home.

The aim is to prove that if *A* = 0, then *A _{q}* =

0 =
2 *A _{q}* cos (18°) + 2

Substitute radical forms for the cosines and multiply both sides by two:

0 =
*A _{q}* √(10 + √20) +

**Elimination of radicals.** Rearrange the previous equation:

− *A _{r}* (√5 + 1) −

At the risk of introducing spurious roots, eliminate some radicals by squaring both sides:

*A _{r}*

Expand, and simplify the radicals:

6 *A _{r}*

Move all radicals to the left-hand member, and divide by two:

√5
(*A _{r}*

Because √5 is irrational, but all of *A _{q}*,

*A _{r}*

− 3

Equate the two left-hand members and simplify:

*A _{r}*

2

**Modulo two.** Filter this last equation into modulo two. First, discard any term that is patently even. Then recall that the product of odd numbers is odd, and the product of anything else is even:

*A _{t}*

Now recall this equation:

*A _{r}*

which in modulo two becomes:

*A _{r}*

*A _{t}* =

*A*,_{q}*A*,_{r}*A*,_{s}*A*are all even._{t}*A*and_{q}*A*are both even, but_{t}*A*and_{r}*A*are both odd._{s}*A*and_{q}*A*are both odd, but_{t}*A*and_{r}*A*are both even._{s}*A*,_{q}*A*,_{r}*A*,_{s}*A*are all odd._{t}

**Modulo four.** Again recall this equation:

*A _{r}*

which in modulo four becomes:

*A _{r}*

Consider the case where *A _{q}* and

1 − 0 − 0 + 1 = 0 [mod 4]

2 = 0 [mod 4] -- **FALSE**

Similarly dismiss the opposite case, where *A _{q}* and

*A*,_{q}*A*,_{r}*A*,_{s}*A*are all even._{t}*A*,_{q}*A*,_{r}*A*,_{s}*A*are all odd._{t}

**Modulo eight.** Note that the square of an odd number is always one more than a multiple of eight. To see this, write the odd number as 2*n* + 1; its square *S* becomes 4*n*^{2} + 4*n* + 1.

- If
*n*is even, write*n*= 2*m*, then*S*= 16*m*^{2}+ 8*m*+ 1 = 8(2*m*^{2}+*m*) + 1. - If
*n*is odd, write*n*= 2*m*+ 1, then*S*= 4(2*m*+ 1)^{2}+ 4(2*m*+ 1) + 1 = 16*m*^{2}+ 24*m*+ 9 = 8(2*m*^{2}+ 3*m*+ 1) + 1

Now consider what happens when all of *A _{q}*,

*A _{r}*

which appears the same in modulo eight:

*A _{r}*

From the comments above pertaining to the square of an odd number, substitute 1 for each of the squares:

1 − 1 − 1 + 1 − 4 *A _{q}*

− 4

Since *A _{q}* and

− 4 (2*B _{q}* + 1) (2

16

Discarding the obvious multiples of eight leaves:

−4 = 0 [mod 8] -- **FALSE**

Now one case is left:

*A*,_{q}*A*,_{r}*A*,_{s}*A*are all even._{t}

**Reduction.** Since all the *A*s are even, then must exist integer *B*s satisfying the equations *A _{q}* = 2

As the existence of solution *A* necessitates the existence of solution *B* = *A* ÷ 2, the existence of solution *B* implies the existence of a solution *C* = *B* ÷ 2 = *A* ÷ 4. From this result, one must conclude that all the components of *A* were multiples of four. By the same token, there must also be solution *D* = *C* ÷ 2 = *B* ÷ 4 = *A* ÷ 8. Now it seems that all the components of *A* were multiples of eight. Further, solution *E* = *D* ÷ 2, which surely exists, calls for the components of *A* to be multiples of sixteen.

This solution-generating process can continue indefinitely, meaning that each component of *A* must have infinitely many factors of two, which cannot happen for a finite nonzero integer. Hence, all the components of a solution to *A* = 0 are themselves zero.